Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Yes, again, this type of question. Similar ones this and this.

I come with another variant. Let $f\in\mathcal{S}$, i.e. Schwartz function, and $g\in L^{p}(\mathbb{R}^d),p\in[1,\infty]$. The following should still hold $$ \partial^\alpha(f*g) = (\partial^\alpha f)*g. $$

Basically we only need to prove the following case and the rest is simply by induction.

Here is my proof for $p\in[1,\infty)$, where are use the proposition $$D_i(f*g) = (D_if)*g,$$ provided $f,g\in\mathcal{S}$ and $D_i:=\frac {\partial}{\partial x_i}$.

Taking arbitrarily fixed $x$, \begin{align*} |D_i(f*g)(x)| &= \left|\lim_{h\rightarrow 0}\frac{f*g(x+he_i)-f*g(x)}{h}\right|\\ &\leq \lim_{h\rightarrow 0}\int_{\mathbb{R}^d}\left|\frac{f(x+he_i-y)-f(x-y)}{h}\right| |g(y)|\,\mathrm{d}y\\ &= \lim_{h\rightarrow 0}\int_{\mathbb{R}^d}|f_h(x-y)| |g(y)|\,\mathrm{d}y\\ &\leq \lim_{h\rightarrow 0}\||f_h(x-\cdot)|\|_{q}\|g\|_p\\ &= \|D_i f\|_{q}\|g\|_p. \end{align*} By taking supremum, we have shown that $D_i(f*g)$ is bounded. Hence $D_i(f*g)(x)$ is well-defined for all $x$.

Then, take a sequence $g_n\in\mathcal{S}$ such that $g_n\rightarrow g$ in $L_p$. Since \begin{align*} |D_i(f*g)-D_i(f*g_n)| &= \lim_{h\rightarrow 0}\left| \int_{\mathbb{R}^d}|f_h(x-y)| |g(y)-g_n(y)|\,\mathrm{d}y \right|\\ &\leq \lim_{h\rightarrow 0}\||f_h(x-\cdot)|\|_{q}\|g-g_n\|_p\\ &= \|D_i f\|_{q}\|g-g_n\|_p \rightarrow \quad \text{as}\quad n\rightarrow 0. \end{align*}

Besides, we know $$D_i (f*g_n) = (D_i f)*g_n,\quad \forall n\in\mathbb{N}.$$ Hence, \begin{align*} |D_i(f*g)-(D_i f)*g| &\leq |D_i(f*g)-(D_i f)*g - D_i (f*g_n) + (D_i f)*g_n|\\ &\leq |D_i(f*g)- D_i (f*g_n)| + |(D_i f)*g_n-(D_i f)*g|\\ &\leq |D_i(f*g)- D_i (f*g_n)| + \|(D_i f)\|_{q}\|g_n-g\|_{p}\rightarrow 0 \quad\text{as}\quad n\rightarrow \infty, \end{align*} where we applied Young's inequality in the last step.

Is it correct? And how should I work on the case $p=\infty$.

share|cite|improve this question
The expression within the $||.||_q$ converges pointwise to $f'(x-y)$. And as you have an integrable majorant ($f$ and $f'$ decay faster than any polynomial), you can apply the dominated convergence theorem. – Vobo Jun 4 '13 at 18:51
@Vobo Thanks for the comment. But we only know $g\in L^p$, how can I handle this? – newbie Jun 4 '13 at 23:42
I am not completely sure about the well-definedness. For instance, $\lim_{n \to \infty} |(-1)^n| \le 1$, hence the sequence is bounded, but the limit does not exist. – Cloudscape Jul 22 '15 at 22:01
up vote 0 down vote accepted

Your answer is correct. Let me show you the case $p=\infty$. As above, it is well-defined. Let $p(x)=(1+||x||^2)^n$. Note that $1/p$ is integrable and $pf$ and $p(D^if)$ are (uniformly) bounded. Then $$ \Big| D^i (f\ast g) (x) - (D^i f\ast g)(x)\Big| \leq ||g||_\infty \lim_{h\to 0} \int |f_h(x-y) - (D^i f)(x-y)| dy. $$ As $f_h\to D^i f$ pointwise, you need to justify the dominated convergence theorem. This can be done by adding the factor $\frac{p(x-y)}{p(x-y)}$ to the integral: You still have $$ \lim_{h\to 0} |p(x-y) f_h(x-y) - p(x-y)(D^i f)(x-y)| = 0 $$ pointwise and this term is uniformly bounded. This bound times $1/p$ is an integrable majorant, so the use of Lebesgue is justified.

share|cite|improve this answer
Thanks. I also figured out that we can use the property of Schwartz function to argue it. – newbie Jun 5 '13 at 21:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.